Re: large integration result for simple problem: 1/x,, also BesselJ

*To*: mathgroup at smc.vnet.net*Subject*: [mg122859] Re: large integration result for simple problem: 1/x,, also BesselJ*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sat, 12 Nov 2011 07:36:44 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201111110955.EAA08514@smc.vnet.net> <7999FCC4-7E77-4ED4-AEF6-AF3CCFAD0FA7@mimuw.edu.pl> <4EBD4CAB.1090505@eecs.berkeley.edu>

Mathematica 8 returns: Integrate[BesselJ[n, b*x], {x, 0, Infinity}, Assumptions -> {Re[n] > -1}] b^(n - 2)*(b^2)^(1/2 - n/2) Andrzej Kozlowski On 11 Nov 2011, at 17:26, Richard Fateman wrote: > On 11/11/2011 7:46 AM, Andrzej Kozlowski wrote: >> Mathematica 8 gives: >> >> Integrate[BesselJ[n, b*x], {x, 0, Infinity}, Assumptions -> {Re[n]> -1, Im[b] == 0}] >> >> Sign[b]^n/Abs[b] > Mathematica 7 gives the same answer with those assumptions. > > What's your point? In my command, where I leave off the Im[b]==0 , I got an answer that includes.. > > If [ Abs[Im[b]] == 0, b^(-2 + n) (b^2)^(1/2 - n/2), ..... > > > To me, Abs[Im[b]] == 0 means the same as Im[b]==0, so why does the answer say Abs[...]==0? > Why is the user required to feed back into the Integrate program the condition it derives, as an Assumption, > in order to get a simplified result? > > Does Mathematica 8 return b^(-2 + n) (b^2)^(1/2 - n/2) ? > > > Does Mathematica 8 do any better for integrating 1/x ? (Sorry, UC Berkeley seems to have let their upgrade lapse). > Wolfram Alpha times out on this. > It also times out on the Bessel integral. > > RJF > > > > > >> >> >> >> Note the additional assumption on b. Without it the result is clearly not true. >> >> Andrzej Kozlowski >> >> >> On 11 Nov 2011, at 10:55, Richard Fateman wrote: >> >>> (at least in Mathematica 7.0) >>> >>> try >>> >>> Integrate[1/x,{x,a,b}] >>> >>> Also >>> >>> Integrate[BesselJ[n, b*x], {x, 0, Infinity}, Assumptions -> Re[n]> -1 ] >>> >>> which returns an expression including this .... >>> >>> b^(-2 + n) (b^2)^(1/2 - n/2) >>> >>> which should be possible to simplify. For example, for b>0, the >>> expression is 1/b. >>> >>> maybe 1/b * If[b>0, 1 , -(-1)^n] or so. >>> >>> I've been playing with integration of expressions involving Bessel >>> functions. Mathematica is sometimes surprising, on both sides of the >>> ledger -- (Yes we Can and No we Can't). >>> >>> >>> RJF >>> >>> >

**References**:**large integration result for simple problem: 1/x,, also BesselJ***From:*Richard Fateman <fateman@cs.berkeley.edu>